$\begin{aligned} &g(x)=\dfrac{8}{4-3x} \\\\ &h(y)=\dfrac{2y-9}{5} \end{aligned}$ $(h\circ g) (4)=$
Solution: Let's start by rewriting $(h\circ g) (4)$ as $h(g(4))$. When evaluating composite functions, we work our way inside out. To evaluate $h(g(4))$, let's first evaluate $g(4)$. Then we'll plug that result into $h$ to find our answer. Let's evaluate $g({4})$. $\begin{aligned}g(x)&=\dfrac{8}{4-3x}\\\\ g({4})&=\dfrac{8}{4-3({4})}~~~~~~~~~~\text{Plug in }x={4}\\\\ &=\dfrac{8}{-8}\\\\ &={-1}\end{aligned}$ We now know that $h(g({4}))$ is the same as $h({-1})$ because $g({4}) = {-1}$. Let's evaluate $h({-1})$. $\begin{aligned}h(y)&=\dfrac{2y-9}{5}\\\\ h({{-1}})&=\dfrac{2({-1})-9}{5}~~~~~~~~~~\text{Plug in }y={-1}\\\\ &=\dfrac{-2-9}{5}\\\\ &=-\dfrac{11}{5}\end{aligned}$ The answer: $(h\circ g)(4) =-\dfrac{11}{5}$